Author Topic: Apparent loss of friction  (Read 617 times)

struktor

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Apparent loss of friction
« on: November 25, 2020, 11:53:11 PM »
 The rope is wrapped around the cylinder and forms a helix.
The force which is pulling the rope acts tangent to the helix, to minimalize lateral slip.
For this purpose, ends of the rope come out at the helix angle.
In the picture 1, β angle is marked.

 helix angle= 90⁰-β

I analyzed the influence of β angle on capstan equation.
Due to above, I came to the conclusion that for this case you can use capstan equation after intruducing an equivalent coefficient friction μᵣ.

 μᵣ=μ*cos(β)

This is consistent with the equation that Alexander Konyukhov⁽⁾ previously introduced for the helix.

It is worth noting that in case where the 30⁰ helix angle is considered, the equivalent friction will be reduced by half.
 μᵣ=μ*cos(β)=μ*cos(90⁰-30⁰)=μ/2
This case was presented in the picture 2.

I am not sure if these conclusions are generally known, so just in case, I have made them aviailable here.


_________________________________________________________________________________________
⁽⁾Alexander Konyukhov, Contact of ropes and orthotropic rough surfaces.(2013)




« Last Edit: November 26, 2020, 12:10:40 AM by struktor »

Dan_Lehman

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Re: Apparent loss of friction
« Reply #1 on: November 26, 2020, 07:42:37 PM »
>> It is worth noting that in case where the 30⁰ helix angle is considered,
>> the equivalent friction will be reduced by half

What significance to knotting do you ascribe to this
asserted reduced friction?
(I'm thinking of "in terms of slide-&-grip friction hitches".)

Thanks,
--dl*
====

struktor

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Re: Apparent loss of friction
« Reply #2 on: November 26, 2020, 09:43:19 PM »
https://en.wikipedia.org/wiki/Capstan_equation
See the table.

Example:
μ=0.4  ; Tl/Th=12
μ=0.2  ; Tl/Th=3.5

12/3.5=3.4

More than 3 times less.

DDK

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Re: Apparent loss of friction
« Reply #3 on: November 28, 2020, 05:28:04 AM »
The Capstan Equation is given as ( see, for example, https://en.wikipedia.org/wiki/Capstan_equation ) :

TLoad = THold* e μφ    ( e represents the Exponential Function )

 and describes the Tension in a rope that is wrapped around a cylinder where:

TLoad :
Tension in the Loaded End of the Rope
THold :
Tension in the Held End of the Rope
μ :
Coefficient of Friction for the Rope/Cylinder Interface
φ :
Total Angle Swept By All Turns of the Rope in Contact with the Cylinder (in radians)

The Tension is at a minimum at the held end of the rope where the rope first makes contact with the cylinder and increases continuously to the maximum Tension in the rope at the loaded end of the rope where the rope loses contact with the cylinder.  The Tension in the rope at any point on the cylinder, TP, can be calculated by inserting into the Capstan Equation the Angle swept from the held point of first contact to the point of interest.  Calling this swept Angle φP, the Tension at point P is given as :

TP = THold* e μφP
« Last Edit: November 28, 2020, 05:30:37 AM by DDK »

DDK

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Re: Apparent loss of friction
« Reply #4 on: November 28, 2020, 05:31:33 AM »
Origins of the Capstan Equation

A rope under Tension lying on a curved surface applies a force per length normal to that surface.  Starting at the loaded end for the sake of discussion, this Normal Force in each small section of the rope is proportional to the Tension at that point and results in a Frictional Force which helps to support the load and reduces the Tension for the next small section of rope.  In this next small section of rope, the Tension and therefore Normal Force AND Frictional Force are a little less than the previous section.  Thus the change (decrease) in Tension is most rapid starting at the loaded end and this decrease continuously slows down as we progress to the held end of the rope.

Some definitions:

T :
Tension in the Rope
N :
Normal Force per Length Applied to the Cylinder Surface
Ff :
Frictional Force Applied to the Rope
Δ :
Signifies a Small Change in or Increment of a Variable
s :
A Length of the Curved Rope ( Arc )
k :
The Curvature of the Rope

The Frictional Force from a small length (section) of rope, Δs, can be calculated by multiplying the Normal Force per Length, N, times this length Δs, times the Coefficient of Friction, μ. In symbols:

Ff = Ns*μ 

Since this Frictional Force from this small section of rope is equivalent to the small decease in the Tension, ΔT, the above can be rewritten as 

ΔT = μ *Ns   ( terms on the right side of the equation were also rearranged )

As mentioned previously, it is the Tension in the rope which results in this Normal Force per Length.  The constant of proportionality is just the Curvature of the Rope, k, lying on this cylindrical surface.  The more tightly curved the surface, the larger the Curvature and the larger the Normal Force for the same applied Tension in the rope.  In symbols:

N = k*T

Putting the last two equations together, eliminating N, produces after rearranging

ΔT = μ *ks*T   Dividing both sides by T results in

ΔT / T  = [ μ *k ]*Δs 

The form of this equation, the Change of a variable divided by that same variable, in this case T, and proportional to a Change in a second variable, in this case s, describes an exponential relationship with the argument of that exponential being the proportionality constant, in this case [ μ *k ], times that second variable, s.  In symbols,

T = T0* e μks        This is the Generalized Form of the Capstan Equation for constant curvature surfaces.

T0 represents the tension at any starting point and s represents the arc length from that same starting point to the point of interest which will have Tension T.

Another example of such a relationship is population.  The increase in the population is proportional to the current population times an increment of time, Δt.  Thus population, P, follows an exponential relationship.

P = P0* e βt    where β is a proportionality constant from

ΔP = [ β ]*Δt*P  or rewritten as   ΔP / P = [ β ]*Δt

« Last Edit: November 29, 2020, 01:35:23 PM by DDK »

DDK

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Re: Apparent loss of friction
« Reply #5 on: November 28, 2020, 05:35:22 AM »
Generalized Capstan Equation Applied to Helical Rope on Cylinders

(1) For rope wrapped perpendicular to the cylinder axis, i.e. no helix

R :
Radius of the Cylinder
β :
Lead Angle of the Helix = 0
φ :
Total Angle Swept By All Turns of the Rope in Contact with the Cylinder (in radians)
s :
Arc Length of the Rope = (R*φ)
k :
The Curvature of the Rope = (1 / R)

Substituting k and s into the Generalized Form of the Capstan Equation for constant curvature surfaces

T = T0* e μks        produces

T = T0* e μ*(1 / R)*(R*φ)        and after combining like terms produces

T = T0* e μφ       which is just the Capstan Equation for no helix.

For cylinders of different Radii, it can be seen that the increase in the Arc Length, s, is exactly offset by the reduction in the Normal Force due to the reduction in the Curvature, k.  Therefore, the Tension is independent of Cylinder Radius.


(2) For rope wrapped at an angle to the cylinder axis, Lead Angle of the Helix ≠ 0

R :
Radius of the Cylinder
β :
Lead Angle of the Helix ≠ 0
φ :
Total Angle Swept By All Turns of the Rope in Contact with the Cylinder (in radians)
s :
Arc Length of the Rope = [R*φ / cos(β)]
k :
The Curvature of the Rope = [cos2(β) / R]

Substituting k and s into the Generalized Form of the Capstan Equation for constant curvature surfaces

T = T0* e μks        produces

T = T0* e μ*[cos2(β) / R]*[R*φ / cos(β)]           and after combining like terms produces

T = T0* e μφcos(β)       which is the Capstan Equation for helical wrapping of a rope.

It can be seen that there is a dependence on the Lead Angle of the Helix because the arc length, s, does not increase as rapidly with the increase in the Lead Angle of the Helix as does the Normal Force and the curvature, k, decrease.  Therefore, the Tension is dependent on the Lead Angle of the Helix.
« Last Edit: November 28, 2020, 06:06:12 AM by DDK »

KC

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Re: Apparent loss of friction
« Reply #6 on: November 28, 2020, 11:03:44 AM »
Am all self taught so some of the Maths hear, jest look like Greek to me!
.
i think that Linears/nonArcs are subject to, but do not rule; cosine & sine the way an Arc does.
i think logically that there are 2 states of loaded rope tensions here : Arc and nonArc(linear).
The linear ropeParts support on the cosine X tension only and only get friction from the sine X tension  expressed towards a host only.
>>only use part of tensions for friction, the sine reflected, diffused part, not the raw,direct cosine part of tensions.
Arcs i think use both sine and cosine for both frictions and support.
>>Arcs use all tensions, including most raw, direct, focused cosine used for frictions in Arcs, in excess of the more diffused sine that give frictions in linears.
But most importantly, the nonArc/linears compound their frictions by distance more intuitively
>>But Arcs compound frictions by degrees of contact>>so larger host gives softer rope bend and spreads out heat and wear
>>But, gives SAME friction in same mated materials over same total degrees of contact; rather counter-intuitively.
My Capstan Math and CoF Reference Sheet
As long as there is contact and force (flow), even if teflon on teflon ;  would have to be friction as a co$t per mated CoFs x pressure x distance/degree increments (linear/radial respectively)?
>>just as electrical resistance per length increments of wire/conducting device in just another type of force flow.
.
Friction free movement would be the Holy Grail of mechanix 'super conductor' ?
>>the whole system built on system of payoffs, nothing free; to balance the whole?
Everything is in Balance for if something is not, it is changing/moving to be so,
>>and in the interim, the process of change itself makes up the imbalance.
Balanced is satiated, not balanced is wandering.
« Last Edit: November 28, 2020, 01:42:12 PM by KC »
"Nature, to be commanded, must be obeyed" -Sir Francis Bacon
We now return you to the safety of normal thinking peoples.
~ Please excuse the interruption; thanx -the mgmt.~

DDK

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Re: Apparent loss of friction
« Reply #7 on: November 28, 2020, 04:05:47 PM »
KC,

Regarding the use of the Greek Alphabet:

" . . . fox hunting! Yes, well, I don't mind that quite so much. At any rate, it's traditional." - George Banks

When one looks very closely at an exceedingly small portion of an arc or anything else, what one sees IS a line.  The closer one looks, the more it looks like a line.  In the limit the length of this portion of the arc goes to zero, it is exactly a line (Thank You Sir Isaac Newton).

For this reason, the Tension in the rope is exactly parallel to the rope at each and every point.  Additionally, this Tension produces a force on the cylinder by the rope which is exactly normal to the surface of the cylinder at each and every point.  This Normal Force produces a Frictional Force which is exactly parallel to the rope at each and every point and therefore, can be directly subtracted from the Sum of Forces (Tensions) for the parallel-to-rope direction at each and every point. This greatly simplifies the calculation because none of the forces need to be broken into their components using sine and cosine. And as you mention, for this reason, the Frictional Force for the arc can be compounded exactly as one does for a line (since it is just a bunch of little line segments).

An important distinction in flat and curved surfaces is the cause of the Frictional Force, that is, from the Normal Force.  For flat surfaces, either weight or applied force is needed.  These can and often may be constant along the length.  Since we are ignoring the weight of the rope, the Normal Force and therefore, the Frictional Force applied by a rope on a curved surface is only dependent on the Tension and Curvature at that point and CHANGES from one point to the next.  For a particular helical Capstan, the Curvature along the spiral is constant, so, it is the Tension that is changing and causing the Frictional Force to change.

The calculation of the Arc Length, s, does require the use of cosine.  To see this, take a sheet of paper and draw a straight line from the lower left corner to the upper right corner.  Now roll the paper into a tube (cylinder) matching two opposing edges.  You now have a perfect helix. Laying the paper flat again, one sees that what was drawn was a right triangle.  The hypotenuse is the Arc Length of the Helix and the bottom of the triangle is the circumference. Using trigonometry, it can be seen that the cosine of the Lead Angle of the Helix equals the circumference divided by the Arc Length (for one pitch of the helix) or

cos(β) = 2R / s     or for only part of the pitch (and only part of the circumference), the part of the circumference is φ*R.  So, in general,

s = φ*R / cos(β)  where s now refers to the Arc Length corresponding to the part of the pitch/circumference.

Similarly, geometrical considerations produce the cosine dependence in the Curvature of the helix.  It is only these geometrical considerations for the Arc Length and Curvature which result in the cosine dependence in the Capstan Equation for the helix.
« Last Edit: January 10, 2021, 03:27:21 PM by DDK »

struktor

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Re: Apparent loss of friction
« Reply #8 on: November 29, 2020, 10:56:11 PM »
Generalized Capstan Equation Applied to Helical Rope on Cylinders

(2) For rope wrapped at an angle to the cylinder axis, Lead Angle of the Helix ≠ 0

β :
Lead Angle of the Helix ≠ 0
φ :
Total Angle Swept By All Turns of the Rope in Contact with the Cylinder (in radians)

T = T0* e μφcos(β)       which is the Capstan Equation for helical wrapping of a rope.
Thank you DDK.

Nomogram to evaluate the influence of the β angle (Lead Angle of the Helix).

Capstan Equation for helical wrapping of a rope.

for β=0
T = T(β=0)* e μφ

for 0<β<90⁰
T = Tβ* e μφcos(β)

After comparing the right sides
Tβ* e μφcos(β) = T(β=0)* e μφ

Further transformations of the equation
Tβ / T(β=0) = e μφ(1-cos(β))

This equation was used to make the nomogram (picture 3)
Tβ / T(β=0) = (e μφ)(1-cos(β))

Extreme values

for β=0   ; Tβ / T(β=0) = 1

for β=90⁰ ; Tβ / T(β=0) = e μφ ; Tβ=T ; (no reduction)

_____________________________________________________________________________________

Nomogram (picture 3) to evaluate the influence of the β angle (Lead Angle of the Helix).


DerekSmith

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Re: Apparent loss of friction
« Reply #9 on: January 25, 2021, 09:43:32 PM »
This seems counter intuitive.  After all, the cord still makes the same number of radians around the static core, so the 'inwards' (i.e. normal) force  is the same, just spread out over a longer distance.  But we have seen from the capstan formulae that circumference is not a factor, so why should the longer path of a spiral have any effect?

Is this theoretical, or has someone actually done the measurement in practice?

DDK

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Re: Apparent loss of friction
« Reply #10 on: January 27, 2021, 05:53:56 PM »
In theory, the Normal Force per Radian is not the same.

The reason that the Capstan does not depend on circumference is because the increase in the path for an increasing radius is exactly offset by a corresponding decrease in the Normal Force per Length resulting in no change in the Normal Force per Radian.  For the helical case, as the helix lead angle or pitch is increased, the path increases more slowly than the Normal Force per Length decreases resulting in the effect that the Normal Force per Radian has decreased.

For example, when the helix angle is increased from zero to 22.5 degrees, this produces an increase in the path of about 8.2%.  The decrease in the Normal Force per Length is about 14.6% resulting in a net decrease of the Normal Force per Radian of 7.6% ( see attached ).

Edited
« Last Edit: January 27, 2021, 08:14:22 PM by DDK »